C. (2, 5) d. (-4, -13)
when you find the value, you SOLVED the equation. you CHECK the equation when you substitute the value in the variables place and check that the equation is true.
6X-5 = 31 (add 5 to both sides ) 6X = 36 ( divide both sides by 6 ) X = 6 ( check ) 6(6) -5 = 31 36-5 = 31 31 = 31
find x when, 2x=128 2x=128 /2 /2 x=128 /2 x=64 then check your answer by doing the following if x=64, then what is 2x? 2(64)=? or 2 * 64=? or 2 x 64=? 2 times 64 equals 128!
x = 2, and y = 0. Here's one way to solve:Multiply the second equation (both sides) by 3, then add the two equations:8x + 6y = 1612x - 6y = 24----------20x = 40 ----> x = 2, then substitute x=2 into one of the original equations:8*2 + 6y = 16 --> 6y = 0, and y=0. Check with the other equation: 4*2 - 2y = 8 ---> -2y = 0, and y = 0.
3x + 4y = 9 is an equation, a statement that states the two expressions 3x + 4y and 9 are equal. We like to know if there are any values of the variables x and y that will make the statement a true one. In our case, we're going to check if x = 2 and y = 5, will satisfy the given equation. In other words, if we substitute the corresponding values of x and y into the equation and the left side will equal to the right side, then (2, 5) is a solution. Otherwise, it's not. Check: 3x + 4y = 9; x = 2 and y = 5 3(2) + 4(5) =? 9 6 + 20 =? 9 26 = 9 False Therefore, (2, 5) is not a solution.
Addiction
x=10 y=4
-11
To check this, you substitute the values for x and y into the equation. At (1,2) x equals 1 and y equals 2 so substituting this in we get: x + 2y = 1 + 2(2) = 1 + 4 = 5 And since it equals 5 like in the original equation the point is represented by it.
If it doesn't have an equal sign, then it's an expression, not an equation. The expression 7x2x is quadratic, because it equals 14x², and something is quadratic if it contains the squared exponent ².
when you find the value, you SOLVED the equation. you CHECK the equation when you substitute the value in the variables place and check that the equation is true.
You can check an answer by substituting it into the equation. x - 17 = 11 25 - 17 = 11 8 = 11 Since this is not true, 25 is not the answer.
32 - 121 - 64 = 36x => -153 = 36x so that x = -4.25
It really depends on the type of equation, but in the simpler cases - those that you are likely to encounter in high school algebra - you will usually need to replace the purported solution into the original equation, then simplify the equation as appropriate. If this results in a true statement (for example, "5 = 5"), then the solution is correct; if you get a false statement (for example, "1 = 0"), then the purported solution is not correct.
An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.
To solve the equation x^2 + 7x - 80, you would get x = 6.105 and x = -13.105, I believe.
I think you are referring to checking a math equation. After you solve an equation you should go back and check your work to make sure you got the right answer. You can do this by plugging your answer back into the equation